text stringlengths 30 4k | source stringlengths 60 201 |
|---|---|
processes 1, 2, 3, solves
BA with 1 possible failure.
• Construct new system S from 2 copies
of A, with initial values as follows:
• What is S?
– A synchronous system of some kind.
– Not required to satisfy any particular
correctness conditions.
3′
1
1
A
3
2
1
2
0
1
S
0
1
0
3
– Not necessarily a correct BA algorithm... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
1-2 from S.
In A, 3 and 1 must agree.
•
• So by indistinguishability, 3 and
1′ agree in S also.
• But we already know that
process 1′ decides 1 and
process 3 decides 0, in S.
• Contradiction!
0
2
0
0
3
1
0
1
S
0
1
3′
1
2′
1′
1
1
1
1
A
0
3
2
Discussion
• We get this contradiction even if the original
algorithm A is... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
process simulates the entire Ii group.
B1
– Bi initializes all processes in Ii with Bi’s initial value.
– At each round, Bi simulates sending messages:
• Local: Just simulate locally.
• Remote: Package and send.
B2
– If any simulated process decides, Bi decides the same (use any).
• Show B satisfies correctness conditi... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
(“if”):
– Suppose both hold.
– Then we can simulate a total-connectivity algorithm.
– Key is to emulate reliable communication from any node i to any
other node j.
– Rely on Menger’s Theorem, which says that a graph is c-connected
(that is, has conn ≥ c) if and only if each pair of nodes is connected
by ≥ c node-di... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
3
0
0
4
1
1
1′
Proof (conn = 2, 1 failure)
• Finally, consider 3′, 4′, and 1 in S:
– Looks to them like they’re in A, with a
faulty process 2.
– In A, they must agree, so they also
agree in S.
– But 3′ decides 0 and 1 decides 1 in S,
contradiction.
• Therefore, we can’t solve BA in
canonical graph, with 1 failure.... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
for 2-Generals problem.
WBA Processor Bounds
• Theorem 4: Weak BA is solvable in an n-node
graph G, tolerating f faults, if and only if n > 3f and
conn(G) > 2f.
• Same bounds as for BA.
• Proof:
– “If”: Follows from results for ordinary BA.
– “Only if”:
• By constructions like those for ordinary BA, but slightly mo... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
Weak BA)
• Now consider a block of 2b + 1 consecutive processes that
begin with 0:
1
2
0
0
• Claims:
3
0
1
0
2
0
3
0
1
0
2
0
3
0
– To all but the endpoints, the execution of S is indistinguishable from
α0, the failure-free execution in which everyone starts with 0, for 1
round.
– To all but two at each end, indisti... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
and the two must have the same decision values.
– Decision values in first and last executions must be the same.
– Contradiction.
Round lower bound, f = 1
• α0: All processes have input 0, no failures.
• …
• αk (last one): All inputs 1, no failures.
• Start the chain from α0.
• Next execution, α1, removes message 1 → ... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
have input 0, no failures, 2 rounds:
– Work toward α n, all 1’s, no failures.
– Each consecutive pair is indistinguishable
0
to some nonfaulty process.
– Use intermediate execs αi, in which:
• Processes 1,…,i have initial value 1.
• Processes i+1,…,n have initial value 0.
• No failures.
0
0
0
Special case: f = 2
• Sh... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
execution where 1 → 2 at round 1, but p2 sends no round 2
messages.
– Now remove the round 1 message 1 → 2.
• Executions look the same to all but 1 and 2 (and they’re nonfaulty).
– Replace all the round 2 messages from p2, one by one, until p2 is
no longer faulty.
• Repeat to remove p1’s round 1 messages to p3, p4,…
... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
is…
• Univalent, if α is either 0-valent or 1-valent (essentially decided).
• Bivalent, if both decisions occur in some extensions (undecided).
Initial bivalence
• Lemma 1: There is some 0-round execution
(vector of initial values) that is bivalent.
• Proof (adapted from [FLP]):
– Assume for contradiction that all 0... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
j1, j2,…jm}.
• Define a chain of (k+1)-round executions,
α0, α1, α2,…, αm.
• Each αl in this sequence is the same as α0
except that i also sends messages to j1,
j2,…jl.
– Adding in messages from i, one at a time.
• Each αl is univalent, by assumption.
• Since α0 is 0-valent, there are 2 possibilities:
– At least one... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
.
And now the final round…
• Lemma 3: There is an f-round execution in which two
nonfaulty processes decide differently.
• Contradicts the problem requirements.
• Proof:
– Use Lemma 2 to get a bivalent (f-1)-round execution α
with ≤ f-1 failures.
– In every 1-round extension of α, everyone who hasn’t
failed must dec... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
0] Stopping agreement
algorithm in which all nonfaulty processes terminate in ≤
min(f′ + 2, f+1) rounds.
– If f′ + 2 ≤ f, decide “early”, within f′ + 2 rounds; in any case decide
within f+1 rounds.
[Keidar, Rajsbaum 02] Lower bound of f′ + 2 for early-
stopping agreement.
– Not just f′ + 1. Early stopping requires an... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
0 and val(α1) = 1.
In the ff extensions of α0 and α1, all nonfaulty processes decide in just
one round.
• Define:
– β0, 1-round extension of α0, in which process i fails, sends only to j.
– β1, 1-round extension of α1, in which process i fails, sends only to j.
• Then:
– β0 looks to j like ff extension of α0, so j dec... | https://ocw.mit.edu/courses/6-852j-distributed-algorithms-fall-2009/0d1d84b9fd1e74bf8e2f8ad916cfca0d_MIT6_852JF09_lec05.pdf |
(cid:2)
2/25/16(cid:2)
2.341J Lecture 6: Extensional Rheometry:
From Entangled Melts to Dilute Solutions
Professor Gareth H. McKinley
Dept. of Mechanical Engineering, M.I.T.
Cambridge, MA 01239
http://web.mit.edu/nnf
1(cid:2)
The Role of Fluid Rheology(cid:1)
•
“Slimy”y
•
”
“Sticky”
3(cid:2)0(cid:2)
• Other manifes... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
excluded from our Creative
Commons license. For more information, see https://ocw.mit.edu/help/faq-fair-use/.
2(cid:2)
Transient Extensional Rheometry(cid:1)
McKinley, (cid:2)
Ann. Rheol. Reviews 2005(cid:2)
• The extensional viscosity is best considered as a transient material
function: follow evolution from equil... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
cid:2)
(e)(cid:2)
(e)(cid:2)
© The British Society of Rheology. All rights reserved. This content is
excluded from our Creative Commons license. For more information,
see https://ocw.mit.edu/help/faq-fair-use/.
© Air Products and Chemicals, Inc. All rights
reserved. This content is excluded from our
Creative Commons li... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
cid:2)
Increasing molecular
alignment at higher strain rates(cid:2)
t
ε(t) = ∫
−∞
ε˙ (t ′ ) dt ′ = ln
L
(t)
L0
Deborah(cid:2)
Number(cid:2)
De = λ(cid:1)ε0
7(cid:2)77
© source unknown. All rights reserved. This content is excluded from our Creative
Commons license. For more information, see https://ocw.mit.edu/help/f... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
)
(cid:1) Mn = 17,000; Mw = 243,000; Mw/ Mn = 14.3
(cid:1) CH3/1000C = 23
(cid:1) Very similar to the IUPAC A reference material
(cid:1) Same polymer as that used by
Münstedt et al., Rheol. Acta 37, 21-29 (1998)
‘Münstedt rheometer’ (end separation method)
SER Principle of Operation(cid:1)
•
•
“Constant Sample Le... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
All rights reserved. This content is excluded
from our Creative Commons license. For more information,
see https://ocw.mit.edu/help/faq-fair-use/.
6(cid:2)
2/25/16(cid:2)
Results for Simple Fluids(cid:1)
• Newtonian Fluids
(cid:1) McKinley & Tripathi, J Rheol., 2000
• Ideal Elastic Fluids
(cid:1) Entov & Hinch, J... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
p
e
s
η
[
λz
101
101
101
100
100
100
106
106
106(cid:2)
107
107
M
w
[g/mol]
M
[g/mol]
Mw [g/mol](cid:2)
w
0.01
0
5
10
15
20
/σ)
R
t/(η0
0
25
30
35
40
Finite Extensibility(cid:1)
•
Important in many biological processes (saliva, spinning of spider silk)
14(cid:2)
© The British Society of Rheology. All rights reserved.... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
Surfactants(cid:2)
• Solutions of EHAC (Schlumberger VES
“clearfrac”,) in brine
Dmid (t) ⇒ (cid:2)ε(t) = − 2
Dmid
dDmid
dt
⇒ η
E,apparent (t)
Yesilata, Clasen & McKinley, JNNFM 133(2) 2006(cid:2)
Kim, Pipe, McKinley; Kor-Aust Rheol. J, 2010(cid:2)
CC by-nc-nd. Some rights reserved. This content is excluded from our Cre... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
York Times Company. All rights reserved. This content is excluded from our Creative
Commons license. For more information, see https://ocw.mit.edu/help/faq-fair-use/.
9(cid:2)
2/25/16(cid:2)
Rayleigh Time Scale
tR ≈
3
ρR0
σ
≈ 0.020s
Break Up of Low Viscosity Liquids(cid:1)
• 0.1 wt% PEO (2x106 g/mol) in water
η0 ≈ 1.... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
cid:24)(cid:31)(cid:19)(cid:28)(cid:31)(cid:24)(cid:32)(cid:36)(cid:1)(cid:7)(cid:6)(cid:10)(cid:6)(cid:38)(cid:12)(cid:1)(cid:9)(cid:28)(cid:20)(cid:21)(cid:25)(cid:1)(cid:1)
The limit of infinite dilution: single molecule experiments(cid:1)
•
Label DNA as a model “supersized” single flexible molecule
The Cross Slot Ap... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
˙ 1
0
Anna et al. J. Rheol. 2001(cid:2)
© The Society of Rheology, Inc. All rights reserved. This content is excluded from our Creative
Commons license. For more information, see https://ocw.mit.edu/help/faq-fair-use/.
(cid:2)
)
y
t
i
s
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c
s
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)
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Increasing (cid... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
R. Cohn, U. Louisville(cid:2)
© The Company of Biologists Ltd.
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excluded from our Creative Commons
license. For more information, see
https://ocw.mit.edu/help/faq-fair-use/.
© source unknown. All
rights reserved. This content
is excluded from our
Creative Commons license.
For ... | https://ocw.mit.edu/courses/2-341j-macromolecular-hydrodynamics-spring-2016/0d326f50e4b153281740a7d4b933e5ea_MIT2_341JS16_Lec06-slides.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.854J / 18.415J Advanced Algorithms
Fall 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
�
�
18.415/6.854 Advanced Algorithms
September 17, 2008
Lecturer: Michel X. Goemans
Lecture 5
Today, we continue the discussion ... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
optimal. In other words
�(f ) = min{� : ∃ potential p : V
→
R such that cp(v, w) ≥ −� for all edges (v, w) ∈ Ef }.
We proved that for all circulations f ,
�(f ) = −µ(f ).
A consequence of this equality is that there exists a potential p such that any minimum mean cost
cycle Γ satisfies cp(v, w) = −�(f ) = µ(f ) fo... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
iterations. Finally, the running time per a single iteration is O(mn) using a variant
of Bellman-Ford (see problem set).
�(f �) ≤ 1 −
�(f ).
�
5-1
1.2 Towards a faster algorithm
In the above algorithm, a significant amount of time is used to compute the minimum cost cycle.
This is unnecessary, as our goal is sim... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
.
(b) Tighten: Update p to p� and � to ��, where p� and �� are chosen such that cp� (v, w) ≥ −��
�
1 �.
for all edges (v, w) ∈ Ef and �� ≤ 1 − n
�
Remark 1 We do not update the potential p every time we push a flow. The potential p gets updated
in the tighten step after possibly several flows are pushed through in... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
are skew-symmetric;
• At least one admissible edge is removed from the residual graph, since we push the maximum
possible amount of flow through the cycle.
5-2
Since we begin with at most m admissible edges, we cannot cancel more than m cycles, as each cycle
canceling reduces the number of admissible edges by at l... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
actually need to find the best possible ��, it is
possible to vastly reduce the running time of the Tighten step to O(n), as follows.
�
When the Cancel step terminates, there are no cycles in the admissible graph Ga = (V, A), the
subgraph of the residual graph with only the admissible edges. This implies that there ... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
.
We consider two cases, depending on whether or not l(v) < l(w).
Case 1: l(v) < l(w). Then
cp� (v, w) = cp(v, w) + (l(w) − l(v))�/n
≥ −� + �/n
= −(1 − 1/n)�.
Case 2: l(v) > l(w), so that (v, w) is not an admissible edge. Then
cp� (v, w) = cp(v, w) + (l(w) − l(v))�/n
≥ 0 − (n − 1)�/n
= −(1 − 1/n)�.
In either ... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
vertex v that has no outgoing edges, then we backtrack, deleting from A the
edges that we backtrack along, until we find an ancestor r of v for which there is another child
to explore. (Notice that every edge we backtrack along cannot be part of any cycle.) Continue
the DFS by exploring paths outgoing from r.
2. If ... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
Time
From the above analysis, we see that the Cancel step requires O(mn) time per iteration, whereas
the Tighten step only requires O(m) time per iteration. In the previous lecture, we determined
that the Cancel-and-Tighten algorithm requires O(min(n log(nC), mn log n)) iterations. Hence the
overall running time is... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
with key[x] = k; if so, returns the
object, and if not, returns false.
• Insert(x): Inserts a new node x into the tree.
• Delete(x): Deletes x from the tree.
• Min: Finds the node with the minimum key from the tree.
• Max: Finds the node with the minimum key from the tree.
• Successor(x): Find the node with the s... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
data structure called splay trees, which is a self-balancing
BST with amortized cost of O(log n) per operation. The idea is that every time a node is accessed,
it gets pushed up to the root of the tree.
The basic operations of a splay tree are rotations. They are illustrated the following diagram.
5-5
ABCxyABCxyzi... | https://ocw.mit.edu/courses/6-854j-advanced-algorithms-fall-2008/0d3338683064d96b5174095829043b93_lec5.pdf |
Lecture 04
Generalization error of SVM.
18.465
Assume we have samples z1 = (x1, y1), . . . , zn = (xn, yn) as well as a new sample zn+1. The classifier trained
on the data z1, . . . , zn is fz1,...,zn .
The error of this classifier is
Error(z1, . . . , zn) = Ezn+1 I(fz1,...,zn (xn+1) =�
yn+1) = Pzn+1 (fz1,...,zn (x... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/0d49e3d6b669cbbb13ef85b0e21357a8_l4.pdf |
-out error. We now prove such a bound for SVMs. Recall that the solution of SVM is
ϕ =
�
n+1 α0
i=1
i yixi.
Theorem 4.1.
L.O.O.E. ≤
min(# support vect., D2/m2)
n + 1
where D is the diameter of a ball containing all xi, i ≤ n + 1 and m is the margin of an optimal hyperplane.
Remarks:
•
•
1
dependence on sam... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/0d49e3d6b669cbbb13ef85b0e21357a8_l4.pdf |
i
=
=
=
�
�
�
α0
i
α0
i (yi(ϕ xi + b) − 1) +
·
��
0
� �
0 − b
αi
α0
i yi
� �� �
0
�
D2
.
m2
�
We now prove Lemma 4.1. Let u ∗ v = K(u, v) be the dot product of u and v, and �u� = (K(u, u))1/2 be
, xn+1 ∈ Rd and y1,
the corresponding L2 norm. Given x1,
· · ·
· · ·
, yn+1 ∈ {−1, +1}, recall th... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/0d49e3d6b669cbbb13ef85b0e21357a8_l4.pdf |
0. Let
αiyi = 0. In other words, α0 corresponds to
, n+1} and α� corresponds to the support vector
αiyixi� 2. Let α0 = argmaxαw(α) subject to αi ≥ 0 and �
αiyi = 0, and ψ = �
αiyixi. Since the
�
�
�
1
2
1
1
classifier trained from {(xi, yi) : i = 1,
· · ·
, p − 1, p + 1,
· · ·
1
↓
· · · ,
,
, n + 1}. Le... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/0d49e3d6b669cbbb13ef85b0e21357a8_l4.pdf |
than the margin m(α0),
7
p ·
�
Lecture 04
Generalization error of SVM.
18.465
and w(α�) ≤ w(α0). On the other hand, the hyperplane determined by α0 − α0 γ might not separate (xi, yi)
p ·
for i =�
p and corresponds to a equivalent or larger “margin” 1/�ψ(α0 − α0 γ)� than m(α�)).
p ·
Let us consider the inequali... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/0d49e3d6b669cbbb13ef85b0e21357a8_l4.pdf |
α� + tγ) − w(α�) =
t
1 (1 − yp · ψ� ∗ xp)2
.
�xp�2
2
For the right hand side,
w(α0 − αp
0 · γ) =
=
�
0 − αp
αi
0 −
�
i − α0
α0
p −
1 �
�
2
�
0 yixi −αp
αi
��
�
ψ0
0 ypxp�2
1
�ψ0�2 + αp
2
�
�2
α0
p
1
2
�xp�2
0 ypψ0 ∗ xp −
�
αp
0�2
1
2
�xp�2
= w(α0) − α0
p(1 − yp · ψ0 ∗ xp) −
1 �... | https://ocw.mit.edu/courses/18-465-topics-in-statistics-statistical-learning-theory-spring-2007/0d49e3d6b669cbbb13ef85b0e21357a8_l4.pdf |
Lecture 14
8.324 Relativistic Quantum Field Theory II
Fall 2010
8.324 Relativistic Quantum Field Theory II
MIT OpenCourseWare Lecture Notes
Hong Liu, Fall 2010
Lecture 14
We now consider the Lagrangian for quantum electrodynamics in terms of renormalized quantities.
L = −
B (γµ(∂µ − ieB AB
µ ) − mB )ψB
= −
− ... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/0d50ef39c801edcfae579fa8d371c85d_MIT8_324F10_Lecture14.pdf |
)ψ
L0 is the free Lagrangian, L1 is the interaction Lagrangian, and Lct is the counter-term Lagrangian. The
parameters Z3 − 1, Z2 − 1 and δm are specified by the following renormalization conditions:
1
iϵ−�(k/) ,
1. For the spinor propagator, S(k) = ik/−m+
k/=−im = 0 (Physical mass condition),
= 0 (Physical field c... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/0d50ef39c801edcfae579fa8d371c85d_MIT8_324F10_Lecture14.pdf |
TEX FUNCTION
Consider the effective vertex we defined before:
Γµ
phys
(k, k) =
µ
≡ −iephysγµ.
k
k
(3)
(4)
This is the physical vertex: it captures the full electromagnetic properties of a spinor interacting with a photon.
As we showed in the previous lecture, the Ward identities impose that
Γµ(k, k) = −ieγµ
... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/0d50ef39c801edcfae579fa8d371c85d_MIT8_324F10_Lecture14.pdf |
the on-shell spinor identities
(7)
ku/ s(k) = −imus(k),
u¯s(k)k/ = −imu¯s(k).
2
̸
Lecture 14
8.324 Relativistic Quantum Field Theory II
Fall 2010
Hence, A, B, and C are scalars, and functions of the scalars k1
the Ward identities, we have that
2 , k2
2 and k1.k2, or, equivalently, of q2 and m. From
and, as ¯... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/0d50ef39c801edcfae579fa8d371c85d_MIT8_324F10_Lecture14.pdf |
k2v − k1v u¯s′ (k2)γv γµus(k1)
k2v − k1v
2
2
k2v + k1v
2
{γµ, γν
} −
)
(
]
u¯s′ (k2)
u¯s′ (k2) [(k2
µ + k1
µ) + iqvσµν ] us(k1).
)
]
[γµ, γν ]
us(k1)
From this, we find that
[
γµF1(q 2) −
iΓµ(k1, k2) = e
]
σµν qν F2(q 2)
.
2m
(9)
�
(10)
F1(q2) and F2(q2) are known as form factors. We have that eF1... | https://ocw.mit.edu/courses/8-324-relativistic-quantum-field-theory-ii-fall-2010/0d50ef39c801edcfae579fa8d371c85d_MIT8_324F10_Lecture14.pdf |
18.600: Lecture 38
Review: practice problems
Scott Sheffield
MIT
1 I Compute the variance of X 2.
I If X1, . . . , Xn are independent copies of X , what is the
probability density function for the smallest of the Xi
Order statistics
I Let X be a uniformly distributed random variable on [−1, 1].
2 I If X1, . . . , X... | https://ocw.mit.edu/courses/18-600-probability-and-random-variables-fall-2019/0db69ebc69650e5313b7624486ab79e4_MIT18_600F19_lec38.pdf |
−1
5 Order statistics
—
answers
I
Var[X 2] = E [X 4] − (E [X 2])2
1
2
x 2dx)2
Z 1
1
5
−
=
1
9
=
4
.
45
Z 1
−1
=
x 4dx − (
1
2
I Note that for x ∈ [−1, 1] we have
Z 1 1
2
P{X > x} =
−1
x
dx
=
1 − x
.
2
If x ∈ [−1, 1], then
P{min{X1, . . . , Xn} > x}
= P{X1 > x, X2 > x, . . . , Xn > ... | https://ocw.mit.edu/courses/18-600-probability-and-random-variables-fall-2019/0db69ebc69650e5313b7624486ab79e4_MIT18_600F19_lec38.pdf |
each
equal to 1 with probability 1/3 and equal to 2 with probability
2/3.
I Compute the entropy H(X ).
9 I Which is larger, H(X + Y ) or H(X , Y )? Would the answer to
this question be the same for any discrete random variables X
and Y ? Explain.
Entropy
I Suppose X and Y are independent random variables, each
eq... | https://ocw.mit.edu/courses/18-600-probability-and-random-variables-fall-2019/0db69ebc69650e5313b7624486ab79e4_MIT18_600F19_lec38.pdf |
, Y ) is larger, and we have H(X , Y ) ≥ H(X + Y ) for any
X and Y . To see why, write a(x, y ) = P{X = x, Y = y } and
b(x, y ) = P{X + Y = x + y }. Then a(x, y ) ≤ b(x, y ) for any
x and y , so
H(X , Y ) = E [− log a(x, y )] ≥ E [− log b(x, y )] = H(X + Y ).
Entropy
—
answers
I
I
1
2
H(X ) = (− log ) +
(− log )... | https://ocw.mit.edu/courses/18-600-probability-and-random-variables-fall-2019/0db69ebc69650e5313b7624486ab79e4_MIT18_600F19_lec38.pdf |
ocw.mit.edu
18.600 Probability and Random Variables
Fall 2019
For information about citing these materials or our Terms of Use, visit:
https://ocw.mit.edu/terms.
15 | https://ocw.mit.edu/courses/18-600-probability-and-random-variables-fall-2019/0db69ebc69650e5313b7624486ab79e4_MIT18_600F19_lec38.pdf |
MIT 2.853/2.854
Manufacturing Systems
Introduction to Simulation
Lecturer: Stanley B. Gershwin
... with many slides by Jeremie Gallien
Copyright c(cid:13)2009 Stanley B. Gershwin.
Copyright c(cid:13)2009 Stanley B. Gershwin.
2
Slide courtesy of Jérémie Gallien. Used with permission.
What is
Simulation?
A computer sim... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
Discrete time
⋆ Discrete event
⋆ Solution of differential equations — I don’t consider
this simulation, but others do.
Copyright c(cid:13)2009 Stanley B. Gershwin.
7
What is
Simulation?
Types of simulation
Copyright c(cid:13)2009 Stanley B. Gershwin.
8
Slide courtesy of Jérémie Gallien. Used with permission.
What is
... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
if (rand(1)<$r1){$alpha1[$step+1]=1}
else{$alpha1[$step+1]=0}}
• if(($alpha1[$step+1]==1)&&($n[$step]<$N)) {$IU=1}
else{$IU=0}
Copyright c(cid:13)2009 Stanley B. Gershwin.
12
Dynamic
Simulation
Discrete Time Simulation
Two-Machine Line
Machine 2 is similar, except:
if(($alpha2[$step+1]==1)&&($n[$step]>0)){$ID=1;$out++... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
clock to the time of the first event on the list.
4. Update the state of the system according to that event.
Copyright c(cid:13)2009 Stanley B. Gershwin.
16
Dynamic
Simulation
Discrete Event Simulation
5. Determine if any new events are made possible by the change
in state, calculate their times (in the same way as in ... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
crete Event Simulation
initialization()
• all variables are initialized
timing(event_list)
• next_eventis determined;
• sim_clock is updated.
event_occurence(next_event)
termination
test?
report_generator(stat_counters)
• system_stateis updated;
• stat_countersis updated;
• rv_generation()is used to
update event_list... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
and the sequence of Zi(ω) satisfies certain
conditions.
Copyright c(cid:13)2009 Stanley B. Gershwin.
24
Dynamic
Simulation
Random numbers
Pseudo-random number generator
Copyright c(cid:13)2009 Stanley B. Gershwin.
25
Slide courtesy of Jérémie Gallien. Used with permission.
Dynamic
Simulation
Random numbers
Pseudo-rand... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
31
Slide courtesy of Jérémie Gallien. Used with permission.
Dynamic
Simulation
Technical Issues
Statistics, Repetitions,
Run Length, and Warmup
• The purpose of simulation is to get quantitative
performance measures such as production rate,
average inventory, etc.
• The data that comes from a simulation is statistical... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
• Required: warmup period ≥ transient period.
• Problem: it is very hard to determine the transient
period.
Copyright c(cid:13)2009 Stanley B. Gershwin.
35
Dynamic
Simulation
Example:
Technical Issues
Statistics, Repetitions,
Run Length, and Warmup
• 20-machine line.
• ri = .1, pi = .02, i = 1, ..., 19; r20 = .1, p20 ... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
000 300000 400000 500000 600000 700000 800000 900000 1e+06
Buffer 1
Buffer 10
Buffer 19
Copyright c(cid:13)2009 Stanley B. Gershwin.
38
Dynamic
Simulation
The Simulation Process
Technical Issues
1 Define the simulation goal
Statistics, Repetitions,
• Never skip!
Run Length, and Warmup
2 Model the system
• Keep the goa... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
simpler.” Albert Einstein.
• The simulation goal should be the guiding light when
deciding what to model
• Start to build your model ON PAPER!
• Get client/user feedback early, and maintain model +
assumption sheet for communication purposes
• Collect data and fit distributions… after modeling the
system, with sensi... | https://ocw.mit.edu/courses/2-854-introduction-to-manufacturing-systems-fall-2016/0dc16a918eabd3f47e3f6d573485dccf_MIT2_854F16_Simulation.pdf |
Lecture 2
(cid:40)(cid:88)(cid:70)(cid:79)(cid:76)(cid:71)(cid:72)(cid:68)(cid:81)(cid:3)(cid:36)(cid:79)(cid:74)(cid:82)(cid:85)(cid:76)(cid:87)(cid:75)(cid:80)(cid:15)(cid:3)(cid:51)(cid:85)(cid:76)(cid:80)(cid:72)(cid:86)
Euclidean gcd Algorithm - Given a, b ∈ Z, not both 0, find (a, b)
• Step 1: If a, b < 0, replace... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/0ddadd3b4c7f386b2ae48a28b6f5ab47_MIT18_781S12_lec2.pdf |
many non-negative
integers less than initial a, there can only be finitely many steps. (Note: because
it decreases by at least 1 at each step, this proof only shows a bound of O(a)
steps, when in fact the algorithm always finishes in time O(log(a)) (left as
(cid:4)
exercise))
To get the linear combination at the same tim... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/0ddadd3b4c7f386b2ae48a28b6f5ab47_MIT18_781S12_lec2.pdf |
.
Lemma 6. If p is prime and p|ab, then p|a or p|b.
Proof. Assume p (cid:45) a, and let g = (p, a). Since p is prime, g = 1 or p, but can’t be
p because g|a and p (cid:45) a, so g = 1. Corollary from last class (4) shows that p|b. (cid:3)
Corollary 7. If p|a1a2 . . . an, then p|ai for some i.
Proof. Obvious if n = 1, a... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/0ddadd3b4c7f386b2ae48a28b6f5ab47_MIT18_781S12_lec2.pdf |
1 . . . q
not
in the set of counterexamples by minimality of n, and so r − 1 = s − 1 and
p2 . . . pr is a permutation of q1 . . . qi 1qi+1 . . . qs, and so r = s and p1p .
2 . . pr is a
−
(cid:4)
permutation of q1q2 . . . qs. (
s. This number is less than n, and
so
−
)
(cid:32)
Theorem 8 (Euclid). There are infinitely m... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/0ddadd3b4c7f386b2ae48a28b6f5ab47_MIT18_781S12_lec2.pdf |
1 +
1
p
1
+
1
2
p1
+
1
p3
1
(cid:18)
(cid:19) (cid:18)
. . .
1 +
+
where
1
1 + +
pi
+
1
p3
2
(cid:19)
1
p2
1
p2
i
1
p2
2
1
p3
i
+
. . .
=
1
−
1
1
pi
<
∞
(cid:19)
(cid:18)
. . . 1 +
. . .
1
pm
+
1
p2
m
(cid:19)
. . .
Since each term is a finite positive number, Σ is also a finite positive number.
After expanding Σ, we can... | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/0ddadd3b4c7f386b2ae48a28b6f5ab47_MIT18_781S12_lec2.pdf |
infinitely many Mersenne primes, ie., primes of the form 2n − 1.
Note: if 2n − 1 is prime, then n itself must be a prime.
4 MIT OpenCourseWare
http://ocw.mit.edu
18.781 Theory of Numbers
Spring 2012
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. | https://ocw.mit.edu/courses/18-781-theory-of-numbers-spring-2012/0ddadd3b4c7f386b2ae48a28b6f5ab47_MIT18_781S12_lec2.pdf |
2.4 DNA structure
×
109 for human, to 9
DNA molecules come in a wide range of length scales, from roughly 50,000 monomers in a
1010 nucleotides in the lily. The latter would be around
λ-phage, 6
thirty meters long if fully stretched. If we consider DNA as a random (non-self avoiding)
Rp, coming to
chain of persistence ... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/0df04225a5ce78faa6d53d8a3167c3ea_MIT8_592JS11_lec16.pdf |
bubble, which is observed experimentally to depend on the length l
of the denatured fragment as
S(l)
≈
bl + c log l + d , with c
1.8kB .
≈
(2.71)
The logarithmic dependence is a consequence of loop closure, and can be justified as follows.
The two single-stranded segments forming a bubble can be regarded as a loop of le... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/0df04225a5ce78faa6d53d8a3167c3ea_MIT8_592JS11_lec16.pdf |
choices by this volume factor. As we
shall see shortly, the parameter g is important in determining the value of the denaturation
temperature, while c controls the nature (sharpness) of the transition.
∝
2.4.1 The Poland–Scheraga model for DNA Denaturation
Strictly speaking, the denaturation of DNA can be regarded as a... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/0df04225a5ce78faa6d53d8a3167c3ea_MIT8_592JS11_lec16.pdf |
polyamino acids,” J. Chem. Phys. 45, 1456 (1966).
49
The single-stranded portions are more flexible, and provide an entropic advantage that is
modeled according to a weight similar to Eqs. (2.73-2.74), as
B(l) =
gl
lc .
(2.77)
Clearly the above weight cannot be valid for strands shorter than a persistence length, but
b... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/0df04225a5ce78faa6d53d8a3167c3ea_MIT8_592JS11_lec16.pdf |
=
Xl=1
zw
zw
1
−
,
while the contribution from a bubble is
∞
∞
B(z) =
zlΩ(l) =
zlgl
lc ≡
f +
c (zg) .
Xl=1
The result for bubbles has been expressed in terms of the functions f +
n (x), frequently en-
countered in describing the ideal Bose gas in the grand canonical ensemble. We recall some
Xl=1
50
(2.82)
(2.83)
∞
X... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/0df04225a5ce78faa6d53d8a3167c3ea_MIT8_592JS11_lec16.pdf |
−
1
zw −
1
−
(cid:20)
f +
c (zg)
,
(cid:21)
(2.88)
from which we can extract physical observables. For example, while the length L is a
random variable in this ensemble, for a given z, its distribution is narrowly peaked around
the expectation value
= z
L
i
h
∂
∂z
log
Z
(z) =
1
zw + f +
1
c−1(zg)
f +
c (zg)
1
zw −
−
.
... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/0df04225a5ce78faa6d53d8a3167c3ea_MIT8_592JS11_lec16.pdf |
Eqs. (2.90-2.91), we see that this limit
is obtained by setting the denominator in these expressions equal to zero, i.e.
from the
condition
f +
c (zg) =
1
zw −
1 .
(2.92)
The type of phase behavior resulting from Eqs. (2.92-2.91), and the very existence of a
transition, depend crucially on the parameter c, and we can d... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/0df04225a5ce78faa6d53d8a3167c3ea_MIT8_592JS11_lec16.pdf |
) in the denominator of Eq. (2.91), we observe that Θ is zero
wc, i.e. the polymer is fully denatured. On approaching the transition point from
≤
52
the other side, Θ goes to zero continuously. Indeed, Eq. (2.93) implies that a small change
1
w
δw
c−1 .
−
≡
Since f +
zg)c−2, we conclude from Eq. (2.91) that the native... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/0df04225a5ce78faa6d53d8a3167c3ea_MIT8_592JS11_lec16.pdf |
c in controlling the nature/existence of the
phase transition can be gleaned by examining the behavior of a single bubble. Examining
the competition between entropy and energy suggests that the probability (weight) of a loop
of length ℓ = 2l is proportional to
p(ℓ)
∝
g
w
ℓ
1
ℓc .
×
(2.95)
(cid:16)
(cid:17)
1.
The proba... | https://ocw.mit.edu/courses/8-592j-statistical-physics-in-biology-spring-2011/0df04225a5ce78faa6d53d8a3167c3ea_MIT8_592JS11_lec16.pdf |
Normal equations
Random processes
6.011, Spring 2018
Lec 15
1
Zillow (founded 2006)
© Zillow. All rights reserved. This content is excluded from our Creative Commons license.
For more information, see https://ocw.mit.edu/help/faq-fair-use/
2 Zestimates
© Zillow. All rights reserved. This co... | https://ocw.mit.edu/courses/6-011-signals-systems-and-inference-spring-2018/0df9fff5e1e92239e0f755a893117b91_MIT6_011S18lec15.pdf |
6.897: Advanced Topics in Cryptography
Lecturer: Ran Canetti
Focus for first half (until Spring Break):
Foundations of cryptographic protocols
Goal: Provide some theoretical foundations of secure
cryptographic protocols:
• General notions of security
• Security-preserving protocol composition
• Some basic construc... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
/4/4): The multi-commitment functionality and
realization. UC Zero Knowledge from UC commitments.
Universal composition with joint state. Problem Set 1 due.
Lecture 10 (3/5/4): Secure realization of any multi-party
functionality with any number of faults (based on
[GMW87,G98,CLOS02]): The semi-honest case. (Static... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
a single framework.
– Conceptual and technical simplicity.
What do we want from a definition of security for a
given task?
• Should be mathematically rigorous
(I.e., should be well-defined how a protocol is modeled
and whether a given protocol is “in” or “out”).
• Should provide an abstraction (“a primitive”) tha... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
n .
– An underlying communication network
• Want to design a “secure” protocol where each p has output
i
f(x1…xn,,r) .That is, want:
– Correctness: The honest parties get the correct function value of the
i
parties’ inputs.
– Secrecy: The corrupted parties learn nothing other than what is
computable from their in... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
Counter example:
Function: F(-,- ) = (r U D,-)
Protocol: P1 chooses r U D, and sends r to P2 .
The protocol is clearly not secret (P2 learns r). Yet, it is possible to generate
P2 ‘s view (it’s a random bit).
(cid:206) Need to consider the outputs of the corrupted parties together with the
outputs of the uncorru... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
ensembles,
i.e. ensembles where each Dk,a is over {0,1}.
• Relations between ensembles:
– Equality: D=D’ if for all k,a, Dk,a = D’k,a .
– Statistical closeness: D~D’ if for all c,d>0 there is a k0
such that for all k> k0 and all a with |a|< kd we have
Prob[xDk,a, x=1] - Prob[xD’k,a, x=1] < k-c .
• Multiparty f... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
• Participants:
– An n-party protocol P=(P1 …Pn). (any n>1)
– Adversary A, controlling a set B of “bad parties” in P.
(ie, the bad parties run code provided by A)
– Environment Z (the initial ITM)
• Computational process:
– Z gets input z
– Z gives A an input a and each good party P an input x
– Until all partie... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
to F
– Bad parties send o F whatever S says. In addition, S sends its own input.
– F evaluates f on the given inputs (tossing coins if necessary) and hands
each party and S its function value. Good parties set their outputs to this
value.
i
i
– S and all parties write their outputs on Z’s subroutine output tape.... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
:
Whatever A computes can be computed
•
given only the prescribed outputs
Input independence: The inputs of the bad parties are
chosen independently of the inputs of the good
parties.
Equivalent formulations:
• Z outputs an arbitrary string (rather than one bit) and Z’s
outputs of the two executions should be ... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
ination.
Is security maintained under these operations?
Examples
• Composition of instances of the same protocol:
– With same inputs/different inputs
– Same parties/different parties/different roles
– Sequential, parallel, concurrent (either coordinated or
uncoordinated).
• “Subroutine composition” (modular compo... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
f.
Notes:
•
In QP, there is at most one protocol active (ie, sending messages) at any
point in time: When P is running, Q is suspended.
It is important that in P all parties terminate the protocol at the same round.
Otherwise the composition theorem does not work…
If P is a protocol in the real-life model then so... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
At the end of the run, output the current state of A.
From the security of P we have that there is an
Sp,Z ~ EXECP,A,Z .
adversary SP such that IDEALf
Note: Here it is important that the input of AP is general
and not only the inputs of the bad parties to the
function.
Adversary H :
– Until the round where the pa... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
A runs from the state in
the output of the external adversary.
– When the internal outputs are generated, hand them to
Z and outputs whatever Z outputs.
Analysis of ZP :
Can verify:
–
–
If the “external system” that ZP interacts with is an ideal
process for f with adversary SP then the simulated Z
sees exactly... | https://ocw.mit.edu/courses/6-897-selected-topics-in-cryptography-spring-2004/0dfd54053823c507a9a117d910d31ff5_lecture1_2.pdf |
MIT OpenCourseWare
http://ocw.mit.edu
6.189 Multicore Programming Primer, January (IAP) 2007
Please use the following citation format:
Saman Amarasinghe, 6.189 Multicore Programming Primer, January
(IAP) 2007. (Massachusetts Institute of Technology: MIT
OpenCourseWare). http://ocw.mit.edu (accessed MM DD, YYYY). ... | https://ocw.mit.edu/courses/6-189-multicore-programming-primer-january-iap-2007/0e56d30fb5cce40a429b63435d2a039a_lec3architctre.pdf |
IAP 2007 MIT
Pipelining Execution
IF: Instruction fetch
EX : Execution
ID : Instruction decode
WB : Write back
Cycles
Instruction #
Instruction i
1
IF
Instruction i+1
Instruction i+2
Instruction i+3
Instruction i+4
2
ID
IF
3
4
5
6
7
8
EX WB
ID
IF
EX WB
ID
IF
EX WB
ID
IF
EX WB
ID
EX WB ... | https://ocw.mit.edu/courses/6-189-multicore-programming-primer-january-iap-2007/0e56d30fb5cce40a429b63435d2a039a_lec3architctre.pdf |
source program
Dependences are a property of programs
Importance of the data dependencies
1) indicates the possibility of a hazard
2) determines order in which results must be calculated
3) sets an upper bound on how much parallelism can possibly
be exploited
● Goal: exploit parallelism by preserving pro... | https://ocw.mit.edu/courses/6-189-multicore-programming-primer-january-iap-2007/0e56d30fb5cce40a429b63435d2a039a_lec3architctre.pdf |
2007 MIT
●
●
Control Dependencies
● Every instruction is control dependent on some set of
branches, and, in general, these control dependencies must
be preserved to preserve program order
if p1 {
S1;
};
if p2 {
S2;
}
● S1 is control dependent on p1, and S2 is control dependent
on p2 but not on p1.
● Control depen... | https://ocw.mit.edu/courses/6-189-multicore-programming-primer-january-iap-2007/0e56d30fb5cce40a429b63435d2a039a_lec3architctre.pdf |
8008
8080
1
'70
Sun's Surface
Rocket Nozzle
Nuclear Reactor
8086
8085
286
386
Hot Plate
486
'80
'90
'00
'10
Image by MIT OpenCourseWare.
Intel Developer Forum, Spring 2004 - Pat Gelsinger
(Pentium at 90 W)
Cube relationship between the cycle time and pow
Prof. Saman Amarasinghe, MIT.
14
6.189 IAP 2007 MIT
Pe... | https://ocw.mit.edu/courses/6-189-multicore-programming-primer-january-iap-2007/0e56d30fb5cce40a429b63435d2a039a_lec3architctre.pdf |
● To maintain throughput T/cycle when each operation has
latency L cycles, need T*L independent operations
● For fixed parallelism:
decreased latency allows increased throughput
decreased throughput allows increased latency tolerance
6.189 IAP 2007 MIT
18
Prof. Saman Amarasinghe, MIT.
e
m
T
i
e
m
T
i
Types... | https://ocw.mit.edu/courses/6-189-multicore-programming-primer-january-iap-2007/0e56d30fb5cce40a429b63435d2a039a_lec3architctre.pdf |
(Transputers, nCube, CM-5)
non-cache-coherent shared memory machines (BBN Butterfly, T3D)
cache-coherent shared memory machines (Sequent, Sun Starfire, SGI
Origin)
● MISD: Multiple Instruction, Single Data
no commercial examples
Prof. Saman Amarasinghe, MIT.
22
6.189 IAP 2007 MIT
My Classification
● By ... | https://ocw.mit.edu/courses/6-189-multicore-programming-primer-january-iap-2007/0e56d30fb5cce40a429b63435d2a039a_lec3architctre.pdf |
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